Làm câu hệ vậy+((
C33.4:
Ta có: \(\left\{{}\begin{matrix}x^2+y^2=1\left(1\right)\\21x+3y+48x^2-48y^2+28xy=69\left(2\right)\end{matrix}\right.\)
Từ pt (1) ta có: \(x^2+y^2=1\Rightarrow y^2=1-x^2\)
Thay vào pt (2) ta được: \(21x+3\sqrt{1-x^2}+48x^2-48\left(1-x^2\right)+28x\sqrt{1-x^2}-69=0\)
\(\Leftrightarrow3\sqrt{\left(1-x\right)\left(1+x\right)}+28x\sqrt{\left(1-x\right)\left(1+x\right)}-21\sqrt{\left(1-x\right)\left(1-x\right)}-48\left(1-x^2\right)-48\left(1-x^2\right)=0\)
\(\Leftrightarrow\sqrt{1-x}\left(3\sqrt{1+x}+28x\sqrt{1+x}-21\sqrt{1-x}-96\left(1+x\right)\sqrt{1-x}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{1-x}=0\\3\sqrt{1+x}+28x\sqrt{1+x}-21\sqrt{1-x}-96\left(1+x\right)\sqrt{1-x}=0\end{matrix}\right.\)
+ Nếu \(\sqrt{1-x}=0\Leftrightarrow1-x=0\Leftrightarrow x=1\Rightarrow y=0\)
+Nếu \(3\sqrt{1+x}+28x\sqrt{1+x}-21\sqrt{1-x}-96\left(1+x\right)\sqrt{1-x}=0\)
\(\Leftrightarrow3\sqrt{1+x}+28x\sqrt{1+x}=21\sqrt{1-x}+96\left(1+x\right)\sqrt{1-x}\)
\(\Leftrightarrow784x^3+952x^2+177x+9=-9216x^3-13248x^2+8775x+13689\)
\(\Leftrightarrow10000x^3+14200x^2-8598x-13680=0\)
\(\Leftrightarrow x=\dfrac{24}{25}\Rightarrow y=\dfrac{7}{25}\)
Thay \(x=\dfrac{24}{24};y=\dfrac{7}{25}\) vào hệ pt ta thấy thoả mãn
\(x=\dfrac{24}{25};y=\dfrac{7}{25}\) là 1 cặp nghiệm của hệ pt
Vậy hệ pt có nghiệm: \(\left(x;y\right)\in\left\{\left(\dfrac{24}{25};\dfrac{7}{25}\right),\left(1;0\right)\right\}\)